Let $M$ be a 3-manifold (the case I am interested is $M$ closed orientable connected hyperbolic); suppose $\pi_1 (M)$ is isomorphic to the fundamental group of a (closed orientable connected) surface (let's say of genus $g \geq 2$). Is $M$ necessarily a product $S \times I$ for some interval $I\subseteq \mathbb{R}$?


If $M$ is a connected closed (i.e. compact without boundary) 3-dimensional manifold, then $\pi_1(M)$ cannot be isomorphic to $\pi_1(S)$, where $S$ is an orientable surface of genus $\ge 2$. You can see this by first noting that $\pi_2(M)=0$ (otherwise, by the sphere theorem, $M$ is a nontrivial connected sum which will imply that $\pi_1(M)$ is a nontrivial free product, which is not the case). Then you observe that $\pi_k(M)=0, k\ge 3$ since the universal cover $\tilde M$ of $M$ is noncompact (here you use that $S$ has positive genus, which implies that the fundamental group is infinite): Thus, $H_k(M)=0$ for all $k\ge 2$. Now, use Hurewicz theorem to conclude that $\pi_k(M)=0$ for all $k\ge 3$.

Thus, $M$ would be homotopy equivalent to $S$ (they have isomorphic fundamental groups and contractible universal covers; now, use Whitehead's theorem). However, $H_3(M)\ne 0$ while $H_3(S)=0$. Contradiction.

What one may ask is about topological classification of compact orientable 3-dimensional manifolds such that $\pi_1(M)\cong \pi_1(S)$. You can still have manifolds of the form: $S\times I$ minus some disjoint open balls. However, such manifolds which are reducible, meaning that there exists a tame 2-sphere in $M$ which does not bound a balls.

It is a theorem (I think, due to Stallings), that if, $M$ is assumed to be oriented, $S$ is oriented, and $M$ is irreducible then, it indeed is homeomorphic to $S\times I$. You can find a proof of this in Hempel's book "3-manifolds". The key is that in this situation there exists a proper homotopy-equivalence $h: M\to S\times I$, i.e., a homotopy equivalence which is a homeomorphism on the boundary. One then proves that such $h$ is homotopic (rel. boundary) to a homeomorphism.

If one allows for connected boundary of $M$, or for nonorientable $M$ or $S$, then the conclusion is that $M$ is homeomorphic to an $I$-bundle over $S$.

Edit. Here is a construction of an open 3-manifold which is homotopy-equivalent to a closed surface $S$ but is not homeomorphic to an interval bundle over $S$. Start with an open contractible 3-dimensional manifold $W$ which is not homeomorphic to $R^3$, say, the Whitehead manifold. Embed (properly and smoothly) a ray $\rho$ into $W$ and then remove a small open tubular neighborhood of $\rho$ from $W$. The result is a contractible manifold with boundary $X$; $\partial X$ is homeomorphic to the open disk $D^2$. Now, take $M=S\times I$, pick an disk $D\subset \partial M$ and glue $M$ to $X$ identifying $\partial X$ and $D$ homeomorphically. Lastly, remove the remaining boundary from the result of gluing. You obtain an open 3-dimensional manifold $N$ which is homotopy-equivalent to $S$ but not homeomorphic to an open interval bundle over $S$. The reason is that $N$ is not "tame". Proving non-tameness of $N$ requires some work to prove (by appealing to its "fundamental group at infinity"); the proof is basically the same as the one showing that $W$ is not tame.

  • $\begingroup$ Are there not nontrivial bundles $I \rightarrow M \rightarrow S$? Or are all such bundles abstractly diffeomorphic to a product? Or all all such $M$ reducible? $\endgroup$ – Jason DeVito Apr 2 '14 at 18:51
  • $\begingroup$ @JasonDeVito: You are right, I forgot about one more assumption, namely that the boundary is not connected. Otherwise, the manifold is an I-bundle. $\endgroup$ – Moishe Kohan Apr 2 '14 at 19:07
  • $\begingroup$ @studiosus thanks for your very complete answer. Can you please tell me something about the non-compact case? I mean: same setting, but $M$ is only required to be boundaryless and orientable (and $I$ will be the whole $\mathbb{R}$). I have checked Hempel's chapter on I-bundles, but he deals only with the compact case. $\endgroup$ – Lor Apr 9 '14 at 14:33
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    $\begingroup$ @Lor: In the case of open manifolds $M$, things work well if $M$ is "tame", i.e., admits a compactification as a manifold with boundary. Otherwise, things are bad since you can use, say Whitehead manifold (contractible and not homeomorphic to $R^3$) to create examples not homeomorphic to line bundles over surfaces. I can provide more details if you care. $\endgroup$ – Moishe Kohan Apr 9 '14 at 16:42
  • $\begingroup$ yes please: I'd really appreciate an explicit construction of an open (non tame?) manifold with fundamental group isomorphic to $\pi_1(S)$ but not homeomorphic to $S\times I$. $\endgroup$ – Lor Apr 11 '14 at 17:11

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